If $X = \frac{B_1 - B_3 + B_2}{\sqrt{2}}$ Where $B_t$ is brownian motion at time $t$.
And I want to find the the distribution of $X$, how would I do so?
$E[X] = 0$ is fairly straight forward.
For variance, however, I run into difficulty in treating:
\begin{equation*} Var(B_1 - B_3 + B_2) = Var( (B_1-B_3) + B_2)) = Var(B_1 - B_3) + Var(B_2) + 2Cov(B_1-B_3,B_2) \end{equation*}
But $Var(B_1 - B_3)$ isn't defined... is it?
Is there something I am missing, another method to calculate the variance of $X$ ? Edit:
\begin{equation*}Var(X) = \frac{1}{2} [ Var(B_1 - B_3 + B_2)] = 0.5 [ Var( (B_1-B_3) + B_2))] = 0.5[Var(B_1 - B_3) + Var(B_2) + 2Cov(B_1-B_3,B_2)] = 0.5 [ Var(B_1) + Var(B_3) + Var(B_2) + 2Cov(B_1-B_3,B_2) -2Cov(B_1,B_3)]= 2 ? \end{equation*}
You have that $X = \frac{B_1 - B_3 + B_2}{\sqrt{2}}$. Since $B$ is a Brownian motion, you know that $B_3-B_2$ and $B_1$ are independent. Thus
$$\text{var}(X) = \frac{1}{2}\text{var}(B_1-B_3+B_2)=\frac{1}{2}\text{var}(B_1-(B_3-B_2))=\frac{1}{2}(\text{var}(B_1)+\text{var}(B_3-B_2)).$$
By definition, the increments $B_s-B_t$ are normally distributed with distribution $N(0,s-t)$ ($s\geq t$), whence $\text{var}(B_s-B_t) = s-t$. Completing the calculation for $\text{var}(X)$, it follows that $$\text{var}(X) = \frac{1}{2}(\text{var}(B_1)+\text{var}(B_3-B_2)) = \frac{1}{2}(1+1)=1.$$
(Recall that $B_1 = B_1-B_0$, and thus $\text{var}(B_1) = 1-0=1$)